TITLE: Efficient visibility blockers in Polya's orchard problem
QUESTION [13 upvotes]: Polya's orchard problem asks for which radius $\rho$ of trees at each lattice point within a distance $R$ of the origin block all lines of sight to the exterior of the orchard.
        


It has been established that rays to infinity are completely blocked iff
$\rho \ge  1/\sqrt{R^2 + 1}$, when $R$ is an integer.
(T.T. Allen, "Polya's orchard problem,"
The American Mathematical Monthly
93(2): 98-104 (1986).)
The above shows a quarter of an orchard with $R=6$, $\rho=1/\sqrt{37}=0.164$,
and some random rays.
I am wondering if disks are the most efficient blockers
in terms of area.  More precisely:

For a given $R$, is there a centrally symmetric convex body $K$
  of area less than $\pi \rho^2$ which when translated to all
  lattice points within distance $R$ of the origin, block all rays from
  the origin to the outside?

My guess is that the answer is Yes, in which case it would
be interesting to know the shape of the area-optimal blockers.
In particular, are they polygons?
The same question may be posed in $\mathbb{R}^d$: are they polytopes?
Edit. Here is the chord construction for $R=2$ from the first paragraph of Douglas's construction, as I understand it:

REPLY [5 votes]: Polya's result can't be improved asymptotically. If we take $\rho\sim\frac cR$ with $c<1$ then positive ratio of rays will not be blocked (as $R\to \infty$). This ratio as a function of $c$ can be calculated explicitly, see for example The Statistics of Particle Trajectories in the Homogeneous
Sinai Problem for a Two-Dimensional Lattice. The lattice $\mathbb{Z}^2$ asymptotically isotropic: this ratio does not depend on angles. In partiqular it means that circles asymptotically are best possible blockers: if in some direction the blocker has smaller width (at least from one side from the center; in this direction from one side the blocker looks like a circle of radius $\rho=cR$ with $c<1$) then the ratio of unblocked rays in this direction will be bigger.
For a fixed $R$ best possible blocker can be founf explicitly. As in original proof we can take all integer points inside given circle and consider all rays passing through this points. If two adjacent rays pass through $A_1=(x_1,y_1)$ and $A_2=(x_2,y_2)$ then blockers at this points must block the ray through $A_3=(x_1+x_2,y_1+y_2)$ (which is outside given circle). Bases of the altitudes from $A_1$ and $A_2$ on $OA_3$ must belong to blockers. It means that optimal blocker is convex hull of all such bases.